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Practice Test

7.EE.2 rewriting expressions to show relationships

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Question 1 of 13

2026 Practice Test D Math, question 2

Question

What is the value of z in x10+y−x5=z10\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}?

  • A1
  • B5
  • Cyy
  • D2y−x2y - x
Show the answer

Correct answer: D

First, multiply both sides of the equation by 10 to eliminate the fractions. In other words, multiply each term by 10.

x10+y−x5=z10\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}

10(x10+y−x5)=10(z10)10\left(\frac{x}{10} + \frac{y - x}{5}\right) = 10\left(\frac{z}{10}\right)

10x10+10(y−x)5=10z10\frac{10x}{10} + \frac{10(y - x)}{5} = \frac{10z}{10}

x+2(y−x)=zx + 2(y - x) = z

Then, distribute and combine like terms to find the value for z.

x+2(y−x)=zx + 2(y - x) = z

x+2y−2x=zx + 2y - 2x = z

2y−x=z2y - x = z

Question analysis

This algebra question gives an equation with three letters and fractions, x over 10 plus (y − x) over 5 equals z over 10, and asks for the value of z. Students clear the denominators, distribute and combine like terms. The two fractions on the left have different denominators, which is where the usual slip happens.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students add and expand linear expressions with fraction coefficients, usually in one letter. Here three letters appear, one numerator is a difference, and the result is an expression rather than a number, which adds abstraction but no new skill.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The equation has three fractions and three letters, and the work is three routine moves: multiply through by 10, distribute, collect the x terms. Most errors come from the middle term, whose denominator 5 differs from the other two; treating it as tenths makes z simply y. Students who test values should avoid x = y, since that pair makes y and 2y − x agree, and x = y = 1 even makes the number 1 fit.

  1. Variables instead of numbers moderate Understanding the question

    The answer is z in terms of x and y, not a number, yet two choices are plain numbers, 1 and 5, which tempt students who expect the equation to give z a single value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Clearing the denominators turns (y − x)/5 into 2(y − x), and the −2x from distributing must then combine with the lone x to leave −x, two places where a factor or a sign slips. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply every term on both sides by 10 so no fractions remain, distribute across the parentheses, and combine the x terms; substituting small numbers for x and y checks the result.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a=b, then 10×a=10×b\text{if } a = b \text{, then } 10 \times a = 10 \times b
a and b are the two sides of an equation. Multiplying both sides by the same number keeps them equal, like adding the same weight to both pans of a balance. Here 10 is a good choice, because it clears away every fraction.
The distributive property
a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c
a is the number outside the parentheses, and b and c are the terms inside. Multiply a by each term inside, and keep the minus sign between them. It works because a groups of (b − c) is a groups of b minus a groups of c.

  1. Sort out the question

    The equation has three letters, x, y and z. You need z written using x and y. The answer will not be one number, because x and y can be anything. The fractions make the work messy, so clear them first.

  2. Multiply every term by 10

    The bottoms are 10, 5 and 10, and each one divides into 10. So multiply every term by 10:

    10×x10+10×y−x5=10×z1010 \times \frac{x}{10} + 10 \times \frac{y - x}{5} = 10 \times \frac{z}{10}

    10 times x/10 is x. Since 10 ÷ 5 = 2, 10 times (y − x)/5 is 2(y − x). And 10 times z/10 is z.

    x+2(y−x)=zx + 2(y - x) = z

  3. Distribute the 2

    Use the distributive property. Multiply the 2 by both y and x:

    2(y−x)=2y−2x2(y - x) = 2y - 2x

    So the equation becomes:

    x+2y−2x=zx + 2y - 2x = z

  4. Combine like terms

    Like terms have the same letter. Here x and −2x are like terms. Since 1 − 2 = −1, they combine to −x:

    x−2x=−xx - 2x = -x

    That leaves:

    2y−x=z2y - x = z

    Watch out: do not add the tops of x/10 and (y − x)/5 straight away. Their bottoms are different. Doing it anyway gives z = y, choice C. Choices A and B are plain numbers, but z changes when x and y do.

    So the answer is D, 2y−x2y - x.

Check your answer

Pick numbers. Let x = 2 and y = 7. The left side is 2/10 + 5/5 = 0.2 + 1 = 1.2. Choice D gives z = 2 × 7 − 2 = 12, and 12/10 = 1.2. Both sides match.

More questions like this

The original DOE question
Question 2 as shown on the DOE’s digital test, page 75 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 75, question 2.

Question 2 of 13

2024 Test A Math, question 7

Question

If 2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x} and x≠0x \neq 0, what expression is represented by □\square?

  • Ayy
  • B2y2y
  • C3y3y
  • D4y4y
Show the answer

Correct answer: C

Use properties of equations to determine the missing expression:

2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

22∙2yx−y2x=□2x\frac{2}{2} \bullet \frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

4y2x−y2x=□2x\frac{4y}{2x} - \frac{y}{2x} = \frac{\square}{2x}

4y−y2x=□2x\frac{4y - y}{2x} = \frac{\square}{2x}

3y2x=□2x\frac{3y}{2x} = \frac{\square}{2x}

The missing expression is 3y3y.

Question analysis

An algebra question with fractions: two fractions with x and 2x in their denominators are subtracted, and students find the numerator that makes the difference equal a fraction over 2x. It tests rewriting a fraction over a common denominator when the denominators contain a variable. Adjusting only part of a fraction is the common mistake.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract fractions with unlike denominators and combine like terms, but rarely see a variable in a denominator. The steps are the same two as with numbers, so the demand is typical once the form is familiar.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 13, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is subtraction of fractions with unlike denominators, written with letters. The box over 2x shows the target denominator, which helps, but each wrong choice matches a familiar fraction error: subtracting the tops before matching the bottoms, or rewriting the first fraction and stopping. Students who rewrite 2y/x as 4y/2x and then subtract y from the new numerator finish in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and there is no number to check unless students substitute values such as x = 1 and y = 2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 2y/x becomes 4y/2x with top and bottom both doubled, while y/2x already has the denominator the box sits over. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite the first fraction over 2x by multiplying its top and bottom by 2, then subtract the numerators, which now share a denominator.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top of the fraction, b is the bottom, and k is any number except zero. Multiplying the top and bottom by the same number does not change the value. It is like multiplying by k over k, which is 1.
Subtracting fractions with the same bottom
ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}
a and b are the tops, and d is the shared bottom, called the denominator. When two fractions have the same bottom, their pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    The box, □\square, stands for a missing expression. An expression is a math phrase made of numbers and letters, like 5y. You need what goes in the box to make the two sides equal. x≠0x \neq 0 means x is not zero, so no fraction divides by zero.

  2. Pick a shared bottom

    You can only subtract fractions when their bottoms match. The bottoms here are x and 2x. The box sits over 2x, so use 2x for both. The second fraction, y2x\frac{y}{2x}, already has it.

  3. Rewrite the first fraction over 2x

    To turn x into 2x, multiply by 2. Multiply the top by 2 as well, so the value stays the same.

    ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}

    2yx=2y×2x×2=4y2x\frac{2y}{x} = \frac{2y \times 2}{x \times 2} = \frac{4y}{2x}

    Watch out: this is only the first fraction. Stopping here and choosing 4y gives choice D.

  4. Subtract the tops

    ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}

    4y2x−y2x=4y−y2x=3y2x\frac{4y}{2x} - \frac{y}{2x} = \frac{4y - y}{2x} = \frac{3y}{2x}

    4y take away y is like 4 apples take away 1 apple. That leaves 3y. Watch out: do not subtract the tops, 2y − y, before the bottoms match. That gives y, choice A.

  5. Match the box

    The left side equals 3y2x\frac{3y}{2x}. The right side is □2x\frac{\square}{2x}. The bottoms are the same, so the tops must be equal. The box is 3y. So the answer is C, 3y3y.

Check your answer

Try numbers. Let x = 1 and y = 2. The left side is 41−22=4−1=3\frac{4}{1} - \frac{2}{2} = 4 - 1 = 3. With 3y in the box, the right side is 3×22×1=62=3\frac{3 \times 2}{2 \times 1} = \frac{6}{2} = 3. Both sides are 3, so 3y works.

More questions like this

The original DOE question
Question 64 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 77, question 64. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 3 of 13

2024 Test A Math, question 25

Question

Points M, N, P and Q, each marked with a dot, lie in that order from left to right on a line. A dimension arrow from M to N is labeled x cm, and a longer dimension arrow from M to P is labeled y cm. No lengths are marked for NP, PQ or MQ.

In the figure above, N is the midpoint of MQ\overline{MQ}. Which segment has length (2x−y)(2x - y) centimeters?

  • APQ\overline{PQ}
  • BNP\overline{NP}
  • CMQ\overline{MQ}
  • DMP\overline{MP}
Show the answer

Correct answer: A

These facts are given:

  • N is the midpoint of MQ\overline{MQ}.
  • MN\overline{MN} has a length of x centimeters.
  • MP\overline{MP} has a length of y centimeters.

Use the first two facts to determine that MQ\overline{MQ} has a length of 2x2x centimeters. So (2x−y)(2x - y) can be represented by MQ−MPMQ - MP, which is PQ\overline{PQ}.

Question analysis

A geometry and algebra question with a labeled figure: points M, N, P and Q lie on a line, MN is x cm, MP is y cm, and N is the midpoint of MQ. Students find which segment has length 2x − y. They must turn the midpoint fact into an expression and read a difference of lengths on the figure.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students evaluate and rewrite expressions but rarely read one as a length on a figure. Combining the midpoint idea from geometry with a difference of expressions, and no numbers, makes this more abstract than typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question reverses the usual task: instead of finding a length, students match an expression to a segment. The midpoint fact turns into MQ = 2x, and taking away the y of MP leaves the piece from P to Q. The wrong choices reward stopping at 2x, reading only the labeled arrows, or matching the letter y; picking x = 5 and y = 7 lets students check with numbers.

  1. Variables instead of numbers moderate Understanding the question

    Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Reading a figure, table or graph moderate Understanding the question

    The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the midpoint to write the length of the whole segment in terms of x, or pick sample values for x and y, then see which segment's length matches 2x − y.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint
MQ=2×MNMQ = 2 \times MN
N is the midpoint of segment MQ, the point exactly in the middle. MN is the half from M to N, and MQ is the whole segment. The whole is two equal halves, so it is 2 times MN.
Adding segments
MP+PQ=MQMP + PQ = MQ
When points lie in order on a line, the short segments add up to the long one. MP runs from M to P, and PQ runs from P to Q. Put end to end, they make MQ.

  1. Sort out the figure

    Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of MQ\overline{MQ}, the point exactly halfway from M to Q. You need the segment that is 2x−y2x - y cm long.

  2. Find the length of MQ

    MN is half of MQ, so MQ is twice as long.

    MQ=2×MN=2xMQ = 2 \times MN = 2x

    MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.

  3. Take MP away from MQ

    Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.

    MP+PQ=MQMP + PQ = MQ

    PQ=MQ−MP=2x−yPQ = MQ - MP = 2x - y

    So PQ is 2x−y2x - y cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is y−xy - x, choice B. So the answer is A, PQ\overline{PQ}.

Check your answer

Try numbers. Say x = 5, so MN = 5 and MQ = 10. Say y = 7, so MP = 7. Then PQ is 10−7=310 - 7 = 3. And 2x−y=10−7=32x - y = 10 - 7 = 3. They match. NP would be 7−5=27 - 5 = 2, which does not.

More questions like this

The original DOE question
Question 82 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 82. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 4 of 13

2023 Test A Math, question 12

Question

If n is an integer and 3n+33n + 3 is an even number, which expression must also represent an even number?

  • A5n+15n + 1
  • B4n+54n + 5
  • C2n+32n + 3
  • Dn+2n + 2
Show the answer

Correct answer: A

The expression 3n+33n + 3 is equal to 3(n+1)3(n + 1), and it is given that this is an even number. The expression (n+1)(n + 1) must also represent an even number, because 3 is an odd number: if (n+1)(n + 1) were odd, 3(n+1)3(n + 1) would be the product of two odd numbers, which is an odd number. If (n+1)(n + 1) is an even number, then n must be an odd number. Choose an odd number to substitute for n in the answer options to find which expression represents an even number:

Let n=3n = 3.

Option A:

5n+1=5(3)+1=165n + 1 = 5(3) + 1 = 16

Option B:

4n+5=4(3)+5=174n + 5 = 4(3) + 5 = 17

Option C:

2n+3=2(3)+3=92n + 3 = 2(3) + 3 = 9

Option D:

n+2=3+2=5n + 2 = 3 + 2 = 5

Since 16 is the only even number, the correct answer is option A.

Question analysis

This number properties question says that 3n + 3 is even for an integer n and asks which of four expressions must also be even. Students reason about odd and even numbers written with a variable. The hidden step is deciding what the given fact says about n itself before testing the expressions.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite and evaluate expressions, but rarely reason about whether an expression is odd or even for every value of a variable. Here that reasoning comes in two layers: first about n, then about each choice.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 23 words.
Corrected from the DOE handbook
  • The explanation gave the wrong reason why n + 1 must be even; the reason was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides its key step: 3n + 3 = 3(n + 1), and since 3 is odd, n + 1 must be even, so n is odd. A student who skips that step and tests a convenient even value never checks the condition and lands on a wrong expression. Once n is known to be odd, any odd test value, such as 1, picks out 5n + 1 at once.

  1. Variables instead of numbers moderate Understanding the question

    No value of n is given; the only fact is that 3n + 3 is even, and it has to be turned into a fact about n itself, that n is odd, before any choice is judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    The parity rules are assumed, not stated: an odd number times an odd number is odd, and an even number plus an odd number is odd. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each expression has to be tested with an odd n or by parity rules; three of the four come out odd whenever n is odd, so none can be dismissed on sight. (Embretson & Wetzel, 1987)

Strategy: First work out what 3n + 3 being even says about n, by factoring out 3 or testing small integers, then check each expression with a value of n that fits.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation said that n + 1 must be even “because 3 is an odd number and the product of an odd number and an even number is an even number.” That fact does not show that n + 1 is even; the reason is that if n + 1 were odd, 3(n + 1) would be the product of two odd numbers, which is odd, and it is given as even. The site now gives that reason. The answer, A, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Pulling out a common factor
a×b+a×c=a(b+c)a \times b + a \times c = a(b + c)
a is a factor, a number being multiplied, that both terms share. b and c are what is left. This is the distributive property run backward. It turns a sum into a product, which is easier to reason about.
Multiplying odd numbers
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}
An even number splits into pairs, like 2, 4 or 6. An odd number has one left over, like 1, 3 or 5. An odd number of odd groups always leaves one extra. So the product is odd.
Adding odd and even numbers
odd+odd=even,even+odd=odd\text{odd} + \text{odd} = \text{even}, \quad \text{even} + \text{odd} = \text{odd}
Two odd numbers each have one left over. Together the two extras make a pair, so the sum is even. An even number plus an odd number still has one left over, so the sum is odd.

  1. Sort out the question

    n is an integer, a whole number that can be positive, negative or 0. You are told that 3n+33n + 3 is even. You need the expression that must be even too. Must means it is even every time. First find out what the fact tells you about n.

  2. Pull out the 3

    Both terms in 3n+33n + 3 have a factor of 3. Pull it out.

    3n+3=3(n+1)3n + 3 = 3(n + 1)

    So 3 times (n+1)(n + 1) is an even number.

  3. Decide whether n is odd or even

    3 is odd. If n+1n + 1 were odd, 3(n+1)3(n + 1) would be odd times odd. That is always odd. But we are told it is even. So n+1n + 1 must be even. One less than an even number is odd. So n is odd. Watch out: don’t test n = 2. Then 3n+3=93n + 3 = 9, which is odd. With n = 2, only n+2n + 2 comes out even, choice D.

  4. Test each choice with an odd n

    Pick an odd number, n = 3. It fits, since 3×3+3=123 \times 3 + 3 = 12 is even.

    • A: 5n+1=5×3+1=165n + 1 = 5 \times 3 + 1 = 16, even
    • B: 4n+5=4×3+5=174n + 5 = 4 \times 3 + 5 = 17, odd
    • C: 2n+3=2×3+3=92n + 3 = 2 \times 3 + 3 = 9, odd
    • D: n+2=3+2=5n + 2 = 3 + 2 = 5, odd

    Only A is even.

  5. Make sure A always works

    Must means every odd n, not just 3. If n is odd, 5n is odd times odd, so it is odd. Odd plus 1 is even. So 5n+15n + 1 is always even. Watch out: 4n is even, but adding the odd 5 makes 4n+54n + 5 odd, choice B. In the same way, 2n+32n + 3 is always odd, choice C. So the answer is A, 5n+15n + 1.

Check your answer

Try another odd number, n = 1. Then 3n+3=63n + 3 = 6, which is even, so n = 1 fits. And 5n+1=65n + 1 = 6 is even too. The other three choices give 9, 5 and 3, all odd. Choice A wins again.

More questions like this

The original DOE question
Question 69 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 69. Open page 78 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 5 of 13

2023 Test A Math, question 29

Question

Integer x is evenly divisible by 3. Which expression below is also evenly divisible by 3?

  • A2x+12x + 1
  • B3x−53x - 5
  • C4x−14x - 1
  • D4x+64x + 6
Show the answer

Correct answer: D

Since integer x is evenly divisible by 3, substitute 3 for x in the answer options to determine which expression is also divisible by 3:

Let x=3x = 3.

Option A:

2x+1=2(3)+1=72x + 1 = 2(3) + 1 = 7

Option B:

3x−5=3(3)−5=43x - 5 = 3(3) - 5 = 4

Option C:

4x−1=4(3)−1=114x - 1 = 4(3) - 1 = 11

Option D:

4x+6=4(3)+6=184x + 6 = 4(3) + 6 = 18

Option D is the only value that is divisible by 3.

Question analysis

This number properties question says an integer x is divisible by 3 and asks which of four linear expressions in x is also divisible by 3, whatever multiple of 3 the integer is. Students can test a multiple of 3 in each expression or think about remainders. Testing a value that is not a multiple of 3 can make a wrong expression look right.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students rewrite and evaluate linear expressions, but they seldom ask whether one is divisible by 3 for every allowed value. One well-chosen substitution settles it, so the demand is about typical for grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 13, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is light; the question turns on choosing a test value that obeys the condition. The habit of trying x = 1 backfires here, because 1 is not a multiple of 3 and it makes two wrong expressions come out divisible. Students who substitute 3, or who see that 4x and 6 are both multiples of 3 whenever x is, settle it quickly, and a second multiple such as 6 confirms it.

  1. Variables instead of numbers moderate Understanding the question

    Nothing pins x down except that it is a multiple of 3, so each expression must be judged for every such x, and any test value has to be a multiple of 3 itself, such as 3 or 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked minor Choosing the answer

    The four expressions differ only in their coefficients and constants, from 2x + 1 to 4x + 6, so none can be dismissed at a glance; each needs a test or a look at its terms. (Embretson & Wetzel, 1987)

Strategy: Substitute a multiple of 3, such as 3 or 6, for x in every expression and keep the one whose value divides evenly by 3; a second multiple confirms it.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding multiples of 3
3a+3b=3(a+b)3a + 3b = 3(a + b)
a and b are whole numbers, so 3a and 3b are multiples of 3. A multiple of 3 is a number that 3 divides evenly, with nothing left over. Adding two multiples of 3 always gives another multiple of 3, because you can pull the 3 out of both parts.

  1. Sort out the question

    “Evenly divisible by 3” means 3 goes into the number with nothing left over. Such a number is called a multiple of 3, like 3, 6, 9 or 12.

    You know that x is a multiple of 3. You need the expression that is a multiple of 3 every time, whatever multiple of 3 x is.

  2. Test x = 3 in every choice

    Pick a number for x that follows the rule. The easiest one is 3. Put it into each expression.

    ChoiceWorkDivisible by 3?
    A2(3)+1=72(3) + 1 = 7No
    B3(3)−5=43(3) - 5 = 4No
    C4(3)−1=114(3) - 1 = 11No
    D4(3)+6=184(3) + 6 = 18Yes

    Watch out: your test number must be a multiple of 3. With x = 1, choice A gives 3 and looks right. That leads to the wrong choice, A.

  3. See why D always works

    One test could be lucky, so see why D works for every x. Since x is a multiple of 3, 4x is one too. For example, 4 × 3 = 12 and 4 × 6 = 24. The number 6 is also a multiple of 3, since 6 = 3 × 2.

    3a+3b=3(a+b)3a + 3b = 3(a + b)

    So 4x + 6 is a multiple of 3 plus a multiple of 3. That is always a multiple of 3.

  4. Rule out the other choices

    Watch out: choice B starts with 3x, but then it subtracts 5. The 5 is not a multiple of 3, so it spoils the result, and 3(3) − 5 = 4. Choices A and C also add or take away a number that is not a multiple of 3.

    So the answer is D, 4x+64x + 6.

Check your answer

Try two more multiples of 3. With x = 6, choice D gives 4(6)+6=304(6) + 6 = 30, and 30 ÷ 3 = 10. With x = 9, it gives 4(9)+6=424(9) + 6 = 42, and 42 ÷ 3 = 14. Both divide evenly, so D keeps working.

More questions like this

The original DOE question
Question 86 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 6 of 13

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

  • A−12
  • B−10
  • C−8
  • D−4
Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

n+(n+1)+(n+2)+(n+3)n+(n+1)+(n+2)+(n+3)

=4n+6=4n+6

Set that expression equal to each of the options and solve for n:

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

n=−184n=-\frac{18}{4} is not an integer.

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

n=−144n=-\frac{14}{4} is not an integer.

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

n=−104n=-\frac{10}{4} is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook
  • A stray word in the explanation’s last sentence was removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

  1. Every choice must be checked moderate Choosing the answer

    The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)

  2. No obvious method minor Working it out

    Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)

  3. Error-prone arithmetic minor Working it out

    Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers
n+(n+1)+(n+2)+(n+3)=4n+6n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.

    You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.

  2. Write the sum with a letter

    You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.

    n+n+n+n+1+2+3=4n+6n + n + n + n + 1 + 2 + 3 = 4n + 6

    Any four consecutive integers add up to 4n + 6.

  3. Test each choice

    Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.

    Choice4n + 64nn
    A−12−18−4.5
    B−10−16−4
    C−8−14−3.5
    D−4−10−2.5

    n must be an integer. Only choice B gives one: n = −4.

  4. Name the four integers

    Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.

    Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.

    So the answer is B, −10.

Check your answer

Add the integers: −4+(−3)+(−2)+(−1)=−10-4 + (-3) + (-2) + (-1) = -10. That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question
Question 86 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 86. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 7 of 13

2023 Test B Math, question 31

Question

If xy=30xy=30, what is the value of 10÷x10\div x in terms of y?

  • A110y\frac{1}{10}y
  • B13y\frac{1}{3}y
  • C3y3y
  • D10y10y
Show the answer

Correct answer: B

Solve xy=30xy=30, the given equation, for x:

xy=30xy=30

x=30yx=\frac{30}{y}

Then:

10÷x=10÷30y=10×y30=13y10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.1 dividing fractions by fractions
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

  1. Variables instead of numbers moderate Understanding the question

    x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction
a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}
a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

  1. Sort out the question

    xyxy means x times y. So x times y is 30. You need 10÷x10 \div x “in terms of y.” That means your answer can use y and numbers, but no x.

  2. Solve for x

    To swap x out, first get x by itself. Divide both sides of xy=30xy = 30 by y.

    x=30yx = \frac{30}{y}

    This says x is 30 divided by y.

  3. Put the fraction in place of x

    Replace x in 10÷x10 \div x with this fraction.

    10÷x=10÷30y10 \div x = 10 \div \frac{30}{y}

    Now use the rule for dividing by a fraction. Flip the fraction and multiply.

    10÷30y=10×y30=10y3010 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}

  4. Simplify the fraction

    Both 10 and 30 can be divided by 10.

    10y30=y3=13y\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y

    Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.

    So the answer is B, 13y\frac{1}{3}y.

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then 10÷x=10÷3=10310 \div x = 10 \div 3 = \frac{10}{3}. Choice B gives 13×10=103\frac{1}{3} \times 10 = \frac{10}{3}. They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question
Question 88 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 8 of 13

2023 Test B Math, question 38

Question

How is 1x−12x\frac{1}{x}-\frac{1}{2x} written as a single fraction? (Assume x≠0x\neq 0.)

  • A12x\frac{1}{2x}
  • B13x\frac{1}{3x}
  • C−1x-\frac{1}{x}
  • D−12x-\frac{1}{2x}
Show the answer

Correct answer: A

1x−12x=22x−12x=12x\frac{1}{x}-\frac{1}{2x}=\frac{2}{2x}-\frac{1}{2x}=\frac{1}{2x}

Question analysis

This algebra question asks students to write 1/x minus 1/(2x) as a single fraction, assuming x is not zero. It tests finding a common denominator when the denominators contain a variable. Getting both fractions over the same denominator before subtracting, and keeping track of the sign, are what matter.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract fractions with unlike denominators and work with linear expressions, but seldom see a variable in a denominator. The steps match those for numerical fractions, so it is a short task once the common denominator is found.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is subtraction of unlike fractions with a letter in the bottom, and every wrong choice is a familiar fraction error: combining the denominators straight across by adding or subtracting, or subtracting the numerators the wrong way round. Substituting x = 1 makes it 1 − 1/2 = 1/2, which rules out the negative choices and 1/3 at once.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and no number checks the result unless students substitute a value such as x = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 1/x becomes 2/2x, and then the subtraction 2 − 1 must be done in the order written, first numerator minus second. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite 1/x with the denominator 2x by multiplying its top and bottom by 2, then subtract the numerators and keep the common denominator.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top, or numerator. b is the bottom, or denominator. k is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, because k over k is 1.
Subtracting fractions with the same bottom
ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
a and b are the tops. c is the shared bottom. When the bottoms match, the pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    You need to write 1x−12x\frac{1}{x} - \frac{1}{2x} as one fraction. The problem says x is not 0. That matters because you can’t divide by 0.

  2. Find a common bottom

    To subtract fractions, their bottoms must match. 2x is 2 times x. So multiply the top and bottom of 1x\frac{1}{x} by 2.

    1x=1×2x×2=22x\frac{1}{x} = \frac{1 \times 2}{x \times 2} = \frac{2}{2x}

    Now both fractions have the bottom 2x.

  3. Subtract the tops

    Subtract the tops in order. Keep the bottom.

    22x−12x=2−12x=12x\frac{2}{2x} - \frac{1}{2x} = \frac{2 - 1}{2x} = \frac{1}{2x}

    Watch out: Never add or subtract the bottoms. Adding them gives 13x\frac{1}{3x}, choice B. Subtracting them gives −1x-\frac{1}{x}, choice C. And keep the order: 2 − 1, not 1 − 2. The wrong order gives choice D.

    So the answer is A, 12x\frac{1}{2x}.

Check your answer

Try x = 1. The problem becomes 1−12=121 - \frac{1}{2} = \frac{1}{2}. Choice A gives 12×1=12\frac{1}{2 \times 1} = \frac{1}{2}. It matches. Choices B, C and D give 13\frac{1}{3}, −1 and −12-\frac{1}{2}.

More questions like this

The original DOE question
Question 95 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 95. Open page 202 of the handbook (PDF)

Question 9 of 13

2022 Test B Math, question 4

Question

The total cost of a book is given by the expression 1.08x1.08x, where x is the price of the book before tax. The tax rate is m%. What is the value of m?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 8

The price of a book before tax is x and after tax is 1.08x1.08x. Subtract to determine the tax rate.

1.08x−x=0.08x1.08x - x = 0.08x, so the tax rate is 0.08.

The tax rate m% can be written as m100\frac{m}{100}.

Solve for m:

m100=0.08\frac{m}{100} = 0.08

100(m100)=100(0.08)100\left(\frac{m}{100}\right) = 100(0.08)

m=8m = 8

Question analysis

The total cost of a book, tax included, is given by the expression 1.08x, where x is the price before tax, and the tax rate is m%. Students find the value of m and type it in. It tests reading a percent increase from an algebraic expression, and the rate has to be entered as a number of percent, not as a decimal.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading a percent increase from an expression such as 1.08x is a named grade 7 idea. It takes one short step, but students must turn a decimal into a percent and type the value with no choices to check.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 3, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression already holds the answer, since 1.08x is the price, 1x, plus a tax of 0.08x. The work lies in pulling the tax apart from the price and then writing it in the form the stem asks for, the number in front of a percent sign. Students who stop at the decimal, or who read the whole 1.08 as 108 percent, grid a value that looks reasonable. Trying a price of $100, which costs $108 with tax, makes the 8 visible.

  1. An answer in an unexpected form moderate Choosing the answer

    The rate is asked for as the m in m%, so the 0.08 found from the expression must become 8, and the grid takes 0.08 or 108 just as readily. (Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    Nothing in the stem says that multiplying a price by 1.08 adds 8% of it; students must split 1.08x into 1x for the price and 0.08x for the tax. (Anderson & Krathwohl, 2001)

  3. Variables instead of numbers minor Understanding the question

    The price is the letter x, never a dollar amount, so the tax has to be read from the coefficient 1.08 rather than worked out from a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Split 1.08x into the price before tax plus the tax added to it, then write the tax part's coefficient as a percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction
m%=m100m\% = \frac{m}{100}
Percent means out of 100. So m% is m hundredths, the fraction m over 100. To change a decimal to a percent, multiply it by 100. To change a percent to a decimal, divide it by 100.

  1. Sort out the question

    x is the price of the book before tax. The total cost with tax is 1.08x, which means 1.08 times x. The tax rate is m%. You need the number m.

    The tax rate is the percent of the price that is added on as tax.

  2. Split the total into price and tax

    The total is the price plus the tax. Split 1.08 into 1 and 0.08.

    1.08x=1x+0.08x=x+0.08x1.08x = 1x + 0.08x = x + 0.08x

    The x part is the price you started with. So the extra 0.08x is the tax. The tax is 0.08 times the price.

  3. Change the decimal to a percent

    The rate is 0.08 as a decimal. Use the formula and set it equal to 0.08.

    m100=0.08\frac{m}{100} = 0.08

    To get m alone, multiply both sides by 100.

    m=0.08×100=8m = 0.08 \times 100 = 8

    So the tax rate is 8%.

  4. Give the answer

    The question asks for m, the number in front of the % sign. That is 8.

    Watch out: 0.08 is the rate as a decimal, one step short. And 1.08, or 108%, counts the price as well as the tax. So you would grid 8.

Check your answer

Try a price of $100. The total is 1.08×100=1081.08 \times 100 = 108 dollars. The tax is 108 − 100 = 8 dollars. And $8 out of $100 is 8%. So m = 8.

More questions like this

The original DOE question
Question 61 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 198, question 61. Open page 198 of the handbook (PDF)

Question 10 of 13

2022 Test B Math, question 51

Question

A clothing store buys shirts for a cost of m dollars each. To determine the selling price of the shirts, the manager increases the purchase cost by 55%. Which expression represents the selling price of each shirt?

  • A0.45m0.45m
  • B0.55m0.55m
  • C1.55m1.55m
  • Dm+0.55m+0.55
Show the answer

Correct answer: C

It is given that m represents the purchase cost.

1m1m represents 100%. The amount the purchase cost was increased is represented by 55%.

Therefore, the selling price will be (100%+55%)m(100\%+55\%)m.

155%=155100=1.55155\%=\frac{155}{100}=1.55

The selling price can be represented by the expression 1.55m1.55m.

Question analysis

A clothing store buys shirts for m dollars each, and the manager sets the selling price by increasing that purchase cost by 55%. Students choose the expression for the selling price of each shirt. It tests writing a percent increase as an algebraic expression, where a percent of an unknown cost has to be handled in letters.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing a percent markup as an expression is a named grade 7 idea and a common classroom task. This is a direct case with one percent and a letter for the cost, with no further steps such as tax.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The markup is an increase by 55%, so the selling price is the whole cost plus 55% of it. The commonest mistake writes only the 55% part, which is the markup and not the price. Others subtract the percent as a discount would, or add 55 cents. Choosing a cost of $100 and finding a $155 price picks out the right expression at once.

  1. Turning words into math moderate Understanding the question

    “Increases the purchase cost by 55%” means the cost plus 55% of it, 100% + 55% = 155% of m, not 55% of m on its own. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    The cost is the letter m, so the markup has to be written as 0.55m and added to m, with no dollar amount to test the expression against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Pick an easy cost such as m = 100 dollars, work out the selling price after a 55% increase, and see which expression gives that price.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new amount=old amount×(1+percent100)\text{new amount} = \text{old amount} \times \left(1 + \frac{\text{percent}}{100}\right)
The old amount is what you start with, here the cost m. The 1 keeps the whole old amount. The percent over 100 adds the increase on top. Use it whenever an amount goes up by a percent.

  1. Sort out the question

    Each shirt costs the store m dollars. The manager raises that cost by 55% to get the selling price. You need an expression for the selling price. An expression is a math phrase with numbers, letters and signs, but no equal sign.

  2. Find the amount of the increase

    The increase is 55% of m. Write 55% as a decimal by dividing by 100.

    55%=55100=0.5555\% = \frac{55}{100} = 0.55

    So the increase is 0.55m dollars, which means 0.55 times m.

    Watch out: 0.55m is only the increase. Stopping here gives choice B.

  3. Add the increase to the cost

    The selling price is the cost plus the increase. Since m is the same as 1m, add the numbers in front.

    m+0.55m=1m+0.55m=1.55mm + 0.55m = 1m + 0.55m = 1.55m

    The percent increase formula gives the same thing:

    new amount=m×(1+55100)=m×1.55=1.55m\text{new amount} = m \times \left(1 + \frac{55}{100}\right) = m \times 1.55 = 1.55m

  4. Give the answer

    Watch out: 0.45m is a 55% decrease, choice A. And m + 0.55 adds only 55 cents, choice D. The increase must grow when the cost grows. So the answer is C, 1.55m1.55m.

Check your answer

Pick a cost of $100, so m = 100. A 55% increase adds $55, for a price of $155. Now try choice C: 1.55×100=1551.55 \times 100 = 155. It matches. Choice B gives only 55, and choice D gives 100.55.

More questions like this

The original DOE question
Question 108 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 212, question 108. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 11 of 13

2018 Test A Math, question 32

Question

The least of 5 consecutive integers is ll, and the greatest is gg. What is the value of l+g2\frac{l+g}{2} in terms of ll?

  • A2l2l
  • B3l3l
  • Cl+2l+2
  • Dl+5l+5
Show the answer

Correct answer: C

The first integer is ll, so the second is l+1l+1, the third is l+2l+2, then l+3l+3, and finally l+4l+4.

Since gg is the fifth and greatest of the integers,

g=l+4g=l+4

Substitute l+4l+4 for gg and simplify:

l+g2=l+l+42=2l+42=l+2\frac{l+g}{2}=\frac{l+l+4}{2}=\frac{2l+4}{2}=l+2

Question analysis

This algebra question about five consecutive integers calls the least one l and the greatest g, then asks for (l + g)/2 written in terms of l. Students must express g using l and simplify a fraction that contains a variable. Counting how far the fifth integer is from the first is where many students slip.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students combine like terms in expressions such as 3(x + 4) − x. This item first asks them to build the expression from a description of consecutive integers, then simplify it, with only letters to work with.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is short once g is written as l + 4, so most errors come from the fence-post count or from testing numbers carelessly. The wrong choices in front of the answer are exactly the ones a single test value cannot rule out, which punishes students who stop at the first match. Using two test values, or simplifying (2l + 4)/2 directly, avoids both traps.

  1. Variables instead of numbers moderate Understanding the question

    Every expression is in l, so the simplification of (l + l + 4)/2 has to be done with letters and cannot be checked against a single number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The description "the least of 5 consecutive integers is l, and the greatest is g" must become g = l + 4, four steps up, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Every choice must be checked minor Choosing the answer

    A single test value fits two choices: l = 1 makes 3l and l + 2 both 3, and l = 2 makes 2l and l + 2 both 4. (Embretson & Wetzel, 1987)

Strategy: Write the integers as l, l + 1, and so on up to the fifth, substitute for g and simplify; then test two different starting values, since one value can match a wrong choice by coincidence.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest of 5 consecutive integers
g=l+4g = l + 4
l is the least (smallest) integer and g is the greatest. Consecutive integers go up by 1 each time. Five integers have only 4 jumps from the first to the last, so you add 4, not 5.
Dividing a sum
a+bc=ac+bc\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. You must divide every part of the top, not just one.

  1. Sort out the words

    Integers are whole numbers, like −2, 0 and 7. Consecutive integers come one right after another, like 8, 9, 10. There are 5 of them here.

    The smallest is called l and the biggest is called g. You must write l+g2\frac{l + g}{2} using only the letter l.

  2. Write all five integers

    Each integer is 1 more than the one before:

    l, l+1, l+2, l+3, l+4l,\ l+1,\ l+2,\ l+3,\ l+4

    The fifth one is the greatest, so g = l + 4. Watch out: five integers have only four jumps between them. Writing g = l + 5 leads to choice D.

  3. Put l + 4 in place of g

    Swap g for l + 4 in the fraction:

    l+g2=l+l+42=2l+42\frac{l + g}{2} = \frac{l + l + 4}{2} = \frac{2l + 4}{2}

    Here l + l is 2l, which means 2 times l.

  4. Split the fraction

    Divide each part of the top by 2:

    2l2+42=l+2\frac{2l}{2} + \frac{4}{2} = l + 2

    So the answer is C, l+2l + 2.

Check your answer

Test two starting numbers. If l = 1, the integers are 1 to 5, and (1 + 5) ÷ 2 = 3, which is 1 + 2. If l = 10, they are 10 to 14, and (10 + 14) ÷ 2 = 12, which is 10 + 2. Use two tests, because l = 1 also makes 3l equal 3, choice B.

More questions like this

The original DOE question
Question 89 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 89. Open page 87 of the handbook (PDF)

Question 12 of 13

2018 Test B Math, question 8

Question

M=3N=P4=Q+5=R7>0M=3N=\frac{P}{4}=Q+5=\frac{R}{7}> 0

Based on the statement above, which variable has the greatest value?

  • AMM
  • BNN
  • CPP
  • DRR
Show the answer

Correct answer: D

Break the equations apart to each equal MM:

M=3NM=3N

M=P4M=\frac{P}{4}

M=Q+5M=Q+5

M=R7M=\frac{R}{7}

Pick a number to substitute into the equations, and solve the equations to find the values of MM, NN, PP, QQ, and RR.

Let M=2M=2. Since all the equations are equal to 2, substitute 2 to find each variable.

M=3NM=3N

2=3N2=3N

23=N\frac{2}{3}=N

M=P4M=\frac{P}{4}

2=P42=\frac{P}{4}

8=P8=P

M=Q+5M=Q+5

2=Q+52=Q+5

−3=Q-3=Q

M=R7M=\frac{R}{7}

2=R72=\frac{R}{7}

14=R14=R

Variable RR has the greatest value.

Question analysis

This algebra question sets five expressions equal to one another and to a positive number, then asks which variable has the greatest value. It tests whether students can rewrite each variable in terms of one shared value and compare them. What makes it tricky is that there are no numbers to compute, so each variable's size has to be reasoned out.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite expressions to see how quantities relate, usually one equation at a time. This chain of five linked expressions, with no numbers to compute and a positivity condition to use, is a reasoning task beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 10, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards a strategy, not a calculation: once the shared value is named, each variable follows in one line and R is clearly largest. Without that step, students judge by how the expressions look, and the look is backwards, since the 3 in 3N makes N small and the 7 under R makes R large. The positivity condition at the end is easy to skip but is what fixes the order.

  1. Variables instead of numbers major Understanding the question

    There are no numbers to compute: five variables are tied together in the one chain M = 3N = P/4 = Q + 5 = R/7, and each must be compared through the shared value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. No obvious method moderate Working it out

    Nothing says how to compare; the efficient move is to name the common value, or pick one such as 12, and rewrite each variable, which students must think of themselves. (Schoenfeld, 1985)

  3. Much to keep track of minor Working it out

    Five relations and the condition > 0 must be held at once; the condition matters, since with a negative common value the order of M, N, P and R would reverse. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Call the shared value k, a positive number, and rewrite each variable as an expression in k; or substitute a convenient positive value such as 12 and compare the results.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Multiplying undoes dividing. So if a divided by b is c, then a is b times c.
Undoing multiplication
b×a=c means a=c÷bb \times a = c \text{ means } a = c \div b
b is the number multiplied by a, and c is the result. Dividing undoes multiplying. So to get a alone, divide c by b.

  1. Understand the chain

    The statement is a chain of equal signs. It says five expressions all have the same value. The last part says that value is greater than 0, so it is positive.

    You need to find which letter has the largest value.

  2. Pick a value to test

    All five expressions are equal, so give them one shared value. Pick a positive number that 3 divides evenly. That way N will be a whole number.

    Let the shared value be 12. Then M=12M = 12, and each of the other four expressions also equals 12.

  3. Solve for each letter

    Undo each operation to find each letter. Divide to undo times. Multiply to undo divide. Subtract to undo plus.

    3N=12, so N=12÷3=43N = 12 \text{, so } N = 12 \div 3 = 4

    P4=12, so P=12×4=48\frac{P}{4} = 12 \text{, so } P = 12 \times 4 = 48

    Q+5=12, so Q=12−5=7Q + 5 = 12 \text{, so } Q = 12 - 5 = 7

    R7=12, so R=12×7=84\frac{R}{7} = 12 \text{, so } R = 12 \times 7 = 84

  4. Compare the values

    The values are M = 12, N = 4, P = 48, Q = 7 and R = 84. R is the largest.

    Watch out: 3N does not make N big. It means 3 times N equals M, so N is only one third of M. That trap points to choice B. Also, P is big, but R is bigger. Check every letter before you pick choice C.

    So the answer is D, R.

Check your answer

Think without numbers. R divided by 7 is M, so R is 7 times M. P is only 4 times M. N is a third of M, and Q is 5 less than M. Since M is positive, 7 times M is the biggest.

More questions like this

The original DOE question
Question 65 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 65. Open page 154 of the handbook (PDF)

Question 13 of 13

2018 Test B Math, question 29

Question

If w−1w-1 is an odd integer, which one of the following must be an even integer?

  • Aw+1w+1
  • B2w−12w-1
  • C2w−22w-2
  • D2w+12w+1
Show the answer

Correct answer: C

One more than an odd integer must be even. One more than w−1w-1 is ww, therefore ww must be even.

Two times an even integer must be even, therefore 2w2w is even. An even integer decreased by 2 must be even.

Therefore, 2w−22w-2 must be even.

Question analysis

This algebra question says that w − 1 is an odd integer and asks which of four expressions must be even. It tests reasoning about odd and even numbers written with a variable. The difficulty is that "must be" means even for every allowed value of w, so one example that happens to work is not enough.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 work evaluates and rewrites expressions but rarely asks whether one is odd or even for every value. The algebra here is light; the demand is reasoning about parity and reading "must be" carefully.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one early inference: since w − 1 is odd, w itself is even. Students who skip that step, or quietly treat w as the odd number, carry the wrong parity into every choice. After that the work is mechanical, because each expression adds an even or an odd amount to an even number. Trying one value, such as w = 4, confirms the choice in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The stem gives no number at all: w − 1 is only known to be odd, so the parity of w itself has to be worked out before any choice can be judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    All four choices are built from w or 2w plus or minus 1 or 2, so each one has to be tested for evenness; three of them come out odd for every allowed w. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    The parity rules are assumed, not stated: one more than an odd number is even, any multiple of 2 is even, and an even number plus or minus 1 is odd. (Anderson & Krathwohl, 2001)

Strategy: Choose a value of w that makes w − 1 odd, evaluate all four expressions, and keep the one that comes out even; then confirm it with a second value.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding 1 to an odd number
odd+1=even\text{odd} + 1 = \text{even}
Odd and even numbers take turns on the number line: 1, 2, 3, 4 and so on. Adding 1 moves you to the next whole number. So the number right after an odd number is always even, and the one after an even number is always odd.
Doubling makes an even number
2×n=even2 \times n = \text{even}
n is any whole number. An even number is one you can split into two equal whole groups. Two times n is always two equal groups of n, so it is always even.

  1. Sort out what you know

    You know that w−1w-1 is an odd integer. An integer is a whole number, like 3, 0 or −5. An even number can be split into two equal groups, like 4 or 10. An odd number is 1 more than an even number, like 3 or 11.

    You need the choice that must be even. Must means it is even for every possible value of ww, not just one.

  2. Decide whether w is odd or even

    The number ww is 1 more than w−1w-1. One more than an odd number is always even. For example, if w−1w-1 is 3, then ww is 4. So ww is even.

    Watch out: do not treat ww itself as the odd number. That mistake makes w+1w+1 look even and leads to choice A.

  3. Show that 2w − 2 is even

    Two times any whole number is even, so 2w2w is even. Taking 2 away from an even number moves you back two spots. You land on another even number. So 2w−22w-2 is even.

    Here is another way to see it. 2w−22w-2 is the same as 2×(w−1)2 \times (w-1). That is 2 times a whole number, so it must be even.

  4. Check the other three choices

    Each one adds or takes away 1 from an even number. That always gives an odd number.

    • A: w+1w+1 is even plus 1, so it is odd.
    • B: 2w−12w-1 is even minus 1, so it is odd.
    • D: 2w+12w+1 is even plus 1, so it is odd.

    Watch out: starting with 2w2w does not make a number even. The +1 in choice D makes it odd.

  5. Give the answer

    Only one choice is even for every possible ww. So the answer is C, 2w−22w-2.

Check your answer

Pick a number. Let w=4w = 4, so w−1=3w-1 = 3, which is odd. Then A is 5, B is 7, C is 6 and D is 9. Only C is even. Try w=10w = 10 too. You get 11, 19, 18 and 21. C is even again.

More questions like this

The original DOE question
Question 86 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 86. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).